# Rolling infinitely many dice

For every integer $n>1$, I have a die $D_n$ with $a_n$ sides labeled $1$ to $a_n$. If $a_n=n^k$, for integer $k>0$, and I roll all the dice at once, what is the probability none of them lands on the side labeled $1$? What is the probability exactly one lands on the side labeled $1$?

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These are not infinite dice. Each one is finite. I think you meant infinitely many dice. It seems to be a widespread error to say "infinite Xs" when what is meant is "infinitely many Xs". Widespread, but still an error. – Michael Hardy Nov 10 '11 at 18:36
I see that's been corrected now. – Michael Hardy Nov 10 '11 at 18:45
These dice are independent? Do you know a formula for events (from a sequence of independent events) all happening? As this is a homework-type problem, I give these hints and not a full solution. – GEdgar Nov 10 '11 at 18:51

First, suppose there are total of $q$ dice.
Given that, for fixed $m$, $D_m = 1$, the probability that no other die rolled at one is: $$\mathbb{P}(\forall_{n \not= m} \, {D_n \not= 1} ; D_m = 1 ) = \frac{1}{1-\frac{1}{a_m}} \prod_{n=2}^q \left( 1- \frac{1}{a_n} \right)$$ The probability that exactly one die lands on 1 is: $$p_k(q) = \sum_{m=2}^q \frac{1}{a_m} \mathbb{P}(\forall_{n \not= m} \, {D_n \not= 1} ; D_m = 1 ) = \prod_{n=2}^q \left( 1- \frac{1}{a_n} \right) \sum_{m=2}^q \frac{1}{a_m-1} = \prod_{n=2}^q \left( 1- \frac{1}{n^k} \right) \cdot \sum_{m=2}^q \frac{1}{m^k-1}$$ For small values of $k$ we have $$p_1(q) = \frac{1}{q} H_{q-1} \qquad p_2(q) = \frac{3}{8} \left(1+\frac{1}{12 q}\right)\left(1-\frac{1}{q} \right)$$ For higher order the probability is expressible in terms of $\Gamma$ functions and poly-gamma functions.
In the limit of infinitely many dice: $$\lim_{q \to \infty} p_1(q) = 0 \qquad \lim_{q \to \infty} p_2(q) = \frac{3}{8} \qquad \lim_{q \to \infty} p_3(q) = \frac{\cosh \left(\frac{\sqrt{3} \pi }{2}\right)}{9 \pi }$$
For each die, the probability of landing on the side labeled one is one over the number of faces of the die, and the probability of landing on a different face is one minus that. Each die is independent, so we can simply multiply the probabilities. The probability that none of them land on the face labeled one is then $$\prod_{n=2}^\infty 1-\frac{1}{n^k}$$ I know no way of solving the general case, but we can have Mathematica calculate the result for various values of $k$, and the results are pretty messy.